Question

The unit of the van der Waals gas equation parameter 'a' in \(\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\) is:

• A. kg m s\(^{-2}\)
• B. atm dm\(^6\) mol\(^{-2}\)
• C. dm\(^3\) mol\(^{-1}\)
• D. kg m s\(^{-1}\)

Hint:

To find the units of constants in physical equations (like 'a' and 'b' in the van der Waals equation), always use the principle of dimensional homogeneity. The pressure correction term \((an^2/V^2)\) must have units of pressure, and the volume correction term \((nb)\) must have units of volume.

Answer 1

The Correct Option is B

To solve the problem of determining the unit of the van der Waals constant 'a' in the gas equation, we start by understanding the given equation:

\(\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\)

This equation is a modification of the ideal gas law to account for real gas behavior. Here, 'a' and 'b' are van der Waals constants:

• 'a' accounts for the intermolecular forces.
• 'b' accounts for the volume occupied by gas molecules.

Our goal is to find the unit of 'a'. Knowing that \(P\) is pressure (units: atm) and \(V\) is volume (units: dm\(^3\)), we can analyze the units of \(\frac{an^2}{V^2}\).

The term \(\frac{an^2}{V^2}\) must have the same units as pressure (atm) for the equation to be dimensionally consistent. Therefore, we have:

Units of \(a\) = Units of pressure × (Units of volume)\(^2\) × (Units of amount of substance)\(^2

Pressure is usually measured in atm, volume in dm\(^3\), and the amount of substance in mol.

Substituting these units, we find:

Units of \(a\) = atm × (dm\(^3\))^2 × mol\(^{-2}\) 

Simplifying, we get:

Units of \(a\) = atm dm\(^6\) mol\(^{-2}\)

The correct option from the given choices is atm dm\(^6\) mol\(^{-2}\).